If $\Omega$ has finite measure and $f \in L^p(\Omega)$ then for $1 \leq q \leq p$ we have $f \in L^q(\Omega)$ and the following estimate.

Question

I was given Hölder's inequality in this form

If $1 \leq p \leq \infty$, $1/p + 1/p' = 1$, $f \in L^p(\Omega)$ and $g \in L^{p'}(\Omega)$ then $fg \in L^1(\Omega)$ and $$||fg||_1 \leq ||f||_p||g||_{p'}$$

and then the text proceeds to saying that if $f \in L^p(\Omega)$, an application of Hölder's inequality to $f = 1\cdot f$ should yield that:

if $\Omega$ has finite measure, for $1 \leq q \leq p$ we have $f \in L^q(\Omega)$ and in particular

$$||f||_q \leq |\Omega|^{1/q - 1/p}||f||_p$$

but I don't really know how to prove such estimate. I can prove $f \in L^q$, though. Applying Hölder to $f = 1\cdot f $ I can only get that

$$||f||_1 \leq |\Omega|^{1 - 1/p}||f||_p$$

This question is similar to this one except that the answers do not prove the estimate, they just present it.

Answer 1

Well you want to estimate $$\int |f|^q d\mu.$$ So we'll follow the hint and write $f = f \cdot 1$. This will give an integrand of the form $|f|^q \cdot 1$, to which we will apply Holder's inequality. We'd like to end up with an integrand of the form $|f|^p$ on the right hand side so apply $\|f^q \cdot 1\|_1 \leq \|f^q\|_\alpha \|1\|_{\alpha'}$ with $\alpha = p/q$. This yields $$\int |f|^q d\mu = \int |f|^q \cdot 1 d\mu \leq \bigg(\int (|f|^q)^{p/q} d\mu\bigg)^{q/p} \bigg( \int 1^{\frac{p}{p-q}} d \mu \bigg)^{\frac{p-q}{p}}$$ by Holder's inequality. This qives $$\|f\|_q \leq \|f\|_p \cdot |\Omega|^{\frac{p-q}{pq}} = |\Omega|^{\frac1q - \frac1p} \|f\|_p$$ as desired.